Question

caitlin won a bag full of money! she has 49 bills in all. she counts $1430. there are twenty dollar bills and fifty dollar bills. how many of each bill does caitlin have? which system best represents the situation?
$\

$$\begin{cases}x + y = 1430 \\\\ 20x + 50y = 49\\end{cases}$$

$
$\

$$\begin{cases}x + y = 49 \\\\ 10x + 5y = 1430\\end{cases}$$

$
$\

$$\begin{cases}x + y = 49 \\\\ 20x + 50y = 1430\\end{cases}$$

$
$\

$$\begin{cases}x + y = 49 \\\\ x + y = 1430\\end{cases}$$

$

Explanation:

Step1: Define Variables

Let \( x \) = number of 20 - dollar bills, \( y \) = number of 50 - dollar bills.

Step2: Total Number of Bills

Total bills: \( x + y = 49 \) (since there are 49 bills in total).

Step3: Total Amount of Money

Total money: \( 20x + 50y = 1430 \) (20 - dollar bills contribute \( 20x \), 50 - dollar bills contribute \( 50y \), total is 1430).
So the system is \(

$$\begin{cases}x + y = 49\\20x + 50y = 1430\end{cases}$$

\)

Answer:

The system \( \boldsymbol{

$$\begin{cases}x + y = 49\\20x + 50y = 1430\end{cases}$$

} \) (the third option with \( x + y = 49 \) and \( 20x + 50y = 1430 \))